8- It is an unexpected answer. 9-I spilled some milk on the counter.
10-It began as a day like any
other day,

Defintion. A subset W of a vector space V is a subspace if

(1) W is non-empty

(2) For every v, ¯ w¯ ∈ W and a, b ∈ F, av¯ + bw¯ ∈ W.

Expressions like av¯ + bw¯, or more generally

X

k

i=1

aiv¯ + i

are called linear combinations. So a non-empty subset of V is a subspace if it is

closed under linear combinations. Much of today’s class will focus on properties of

subsets and subspaces detected by various conditions on linear combinations.

Theorem. If W is a subspace of V , then W is a vector space over F with operations

coming from those of V .

In particular, since all of those axioms are satisfied for V , then they are for W.

We only have to check closure!

Examples:

Defintion. Let F

n = {(a1, . . . , an)|ai ∈ F} with coordinate-wise addition and scalar

multiplication.

This gives us a few examples. Let W ⊂ F

n be those points which are zero except

in the first coordinate:

W = {(a, 0, . . . , 0)} ⊂ F

n

.

Then W is a subspace, since

a · (α, 0, . . . , 0) + b · (β, 0, . . . , 0) = (aα + bβ, 0, . . . , 0) ∈ W.

If F = R, then W0 = {(a1, . . . , an)|ai ≥ 0} is not a subspace. It’s closed under

addition, but not scalar multiplication.

We have a number of ways to build new subspaces from old.

Proposition. If Wi for i ∈ I is a collection of subspaces of V , then

W =

\

i∈I

Wi = {w¯ ∈ V |w¯ ∈ Wi∀i ∈ I}

is a subspace.

Proof. Let ¯v, w¯ ∈ W. Then for all i ∈ I, ¯v, w¯ ∈ Wi

, by definition. Since each Wi

is

a subspace, we then learn that for all a, b ∈ F,

av¯ + bw¯ ∈ Wi

,

and hence av¯ + bw¯ ∈ W. ¤

Thought question: Why is this never empty?

The union is a little trickier.

Proposition. W1 ∪ W2 is a subspace iff W1 ⊂ W2 or W2 ⊂ W1.

i hope this helped have a nice day/night :)